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timurjin [86]
3 years ago
10

Help in this question please?

Mathematics
1 answer:
Artyom0805 [142]3 years ago
4 0

my issues is how we finna see this it's small but I think that it could be 25 or 7

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Identify the pattern then write the next three terms in each sequence 2,8,32,128​
Allisa [31]

Step-by-step explanation:

the difference is the previous term multiplied by 4 to get the next term

2,8,32,128,512,2048,8192

Hope that helps :)

3 0
3 years ago
Steven works at the Pi Day food truck and he gets paid $9 per hour x. If 1 point
vitfil [10]
P(x) =9x +30

the tips are a flat rate so it’s a constant 30 while the 9 per hour is dependent on the hours worked which is why it is attached to x as a coefficient
8 0
3 years ago
What is the distance of this line?<br><br>Which is the right answer? ​
Shtirlitz [24]

Answer:

Step-by-step explanation:

We can use the distance formula derived from the Pythagorean theorem

D = \sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}

the two points given are

(0, 3) and (-2, -3)

(x_2-x_1) = (0-(-2)) = 2\\(y_2-y_1) = (3-(-3)) = 6\\D = \sqrt{(2)^2 + (6)^2} \\D = \sqrt{4 + 36} \\D = \sqrt{40} = 6.324

6 0
3 years ago
How many steps are there in the problem-solving process
Alex73 [517]

Answer:

Step-by-step explanation: There are four steps

8 0
3 years ago
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

3 0
3 years ago
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